Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $a = \dfrac{k + 5}{5k^2 - 30k} \div \dfrac{-3k - 15}{k^2 - 16k + 60} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{k + 5}{5k^2 - 30k} \times \dfrac{k^2 - 16k + 60}{-3k - 15} $ First factor the quadratic. $a = \dfrac{k + 5}{5k^2 - 30k} \times \dfrac{(k - 6)(k - 10)}{-3k - 15} $ Then factor out any other terms. $a = \dfrac{k + 5}{5k(k - 6)} \times \dfrac{(k - 6)(k - 10)}{-3(k + 5)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (k + 5) \times (k - 6)(k - 10) } { 5k(k - 6) \times -3(k + 5) } $ $a = \dfrac{ (k + 5)(k - 6)(k - 10)}{ -15k(k - 6)(k + 5)} $ Notice that $(k + 5)$ and $(k - 6)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ \cancel{(k + 5)}(k - 6)(k - 10)}{ -15k\cancel{(k - 6)}(k + 5)} $ We are dividing by $k - 6$ , so $k - 6 \neq 0$ Therefore, $k \neq 6$ $a = \dfrac{ \cancel{(k + 5)}\cancel{(k - 6)}(k - 10)}{ -15k\cancel{(k - 6)}\cancel{(k + 5)}} $ We are dividing by $k + 5$ , so $k + 5 \neq 0$ Therefore, $k \neq -5$ $a = \dfrac{k - 10}{-15k} $ $a = \dfrac{-(k - 10)}{15k} ; \space k \neq 6 ; \space k \neq -5 $